Question

Let where a , BE R such that aメβ. Select an invertible matrix P such that P- AP a diagonal matrix Hence, select all the true statements from the list below: 12 12 12 13 13 13 15 15 18 18 18

Answer 1

in A = PDP^-1form P iseigenvector matrix,and D is eigenvalues Eigenvalue and eigenvectors calculations det (A - I) = ab = 0 eigenvalue: a = 0 and b = 0 eigenvalue -a: A - aI = [a - a 0 1 b - a] [x y] = [0 0] solution y = 0, x = tso [1 0] is eigenvector eigenvalue -b: A - bI = [a - b 0 1 b - b] [x y = [0 0] solution y = t, x = -t/a - b so [-1/a - b 1] = [1 b - a] is eigenvector P = [1 0 1 b - a] so option 3rd is correct part [2] \r\n D = [a 0 0 b], [1 0 1/a - b -1/a - b] A = PDP^-1 = [1 0 1 b -a] [a 0 0 b] [1 0 1/a - b -1/a - b] A^n = PD^n P^-1 = [1 0 1 b - a] [a^n 0 0 b^n] [1 0 1/a - b -1/a - b] A^n = [a^n 0 a^n - b^n/a - b 0 b^n]

so option 3rd is correct.

Part [2]

$\\D = \begin{bmatrix} a &0 \\ 0 & b \end{bmatrix},\ \begin{bmatrix} 1 &\frac{1}{a-b} \\ 0& \frac{-1}{a-b} \end{bmatrix}\\ \\A = PDP^{-1} = \begin{bmatrix} 1& 1\\0 &b-a \end{bmatrix}\begin{bmatrix} a &0 \\ 0 & b \end{bmatrix}\begin{bmatrix} 1 &\frac{1}{a-b} \\ 0& \frac{-1}{a-b} \end{bmatrix}\\ \\A^{n} = PD^nP^{-1} =\begin{bmatrix} 1& 1\\0 &b-a \end{bmatrix}\begin{bmatrix} a^n &0 \\ 0 & b^n \end{bmatrix}\begin{bmatrix} 1 &\frac{1}{a-b} \\ 0& \frac{-1}{a-b} \end{bmatrix}\\ \\A^{n} = \begin{bmatrix} a^n &\frac{a^n-b^n}{a-b} \\ 0& b^n \end{bmatrix}$

Option 1st and 3rd are correct.

It was a pretty long question and with multiple parts. As per Chegg instruction, I could have answered only one part and moved on. But I went on to answer all of them in hope that you will upvote the answer. Please upvote the answer